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Stepping Numbers CodeChef Solution
Problem -Stepping Numbers CodeChef Solution
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Stepping Numbers CodeChef Solution in C++17
“
#include <stdio.h>
#include <stdlib.h>
#define ullint unsigned long long
#define uint unsigned int
#define MOD 4294967143ULL
#define MODINV 1789569643ULL
typedef struct {
ullint a;
ullint b;
} surd;
inline ullint modprod(ullint a, ullint b) {
return (a*b) % MOD;
}
inline ullint modminus(ullint a, ullint b) {
a = a % MOD;
b = b % MOD;
if (a >= b)
return a - b;
else
return (MOD - b) + a;
}
inline surd mult(surd s1, surd s2) {
surd res;
res.a = (modprod(s1.a,s2.a) + modprod(modprod(s1.b,s2.b),3)) % MOD;
res.b = (modprod(s1.a,s2.b) + modprod(s1.b,s2.a)) % MOD;
return res;
}
ullint p3array[65];
ullint p2array[65];
surd surdarray[65];
void init() {
p2array[1] = 2;
p3array[1] = 3;
surdarray[1].a = 2;
surdarray[1].b = 1;
for(int i = 2; i < 65; i++) {
p2array[i] = modprod( p2array[i-1] , p2array[i-1] );
p3array[i] = modprod( p3array[i-1] , p3array[i-1] );
surdarray[i] = mult(surdarray[i-1],surdarray[i-1]);
}
}
void solve(ullint a) {
if (a == 0 || a % 2 == 1) {
printf("0\n");
return;
}
a = a >> 1;
ullint p2 = 1;
ullint p3 = 1;
surd ps;
ps.a = 1;
ps.b = 0;
int i = 1;
while(a > 0) {
if ( (a & 1) == 1) {
p2 = modprod(p2, p2array[i]);
p3 = modprod(p3, p3array[i]);
ps = mult(ps,surdarray[i]);
}
i++;
a = a >> 1;
}
surd psminus;
psminus.a = modminus(2*ps.a, 3*ps.b);
psminus.b = modminus(2*ps.b, 3*ps.a);
ullint res1 = modprod(MODINV, (2*psminus.a + p3 + 2*p2 + 3) % MOD );
ullint res2 = 2*(2*ps.a + p3 + p2 + 1);
ullint res = modminus(res2, res1);
printf("%llu\n", res);
}
main() {
int T;
scanf("%d", &T);
int i;
init();
for(i = 0; i < T; i++) {
ullint a;
scanf("%llu", &a);
solve(a);
}
}
Stepping Numbers CodeChef Solution in C++14
“#include <stdio.h>
#include <stdlib.h>
#define ullint unsigned long long
#define uint unsigned int
#define MOD 4294967143ULL
#define MODINV 1789569643ULL
typedef struct {
ullint a;
ullint b;
} surd;
inline ullint modprod(ullint a, ullint b) {
return (a*b) % MOD;
}
inline ullint modminus(ullint a, ullint b) {
a = a % MOD;
b = b % MOD;
if (a >= b)
return a - b;
else
return (MOD - b) + a;
}
inline surd mult(surd s1, surd s2) {
surd res;
res.a = (modprod(s1.a,s2.a) + modprod(modprod(s1.b,s2.b),3)) % MOD;
res.b = (modprod(s1.a,s2.b) + modprod(s1.b,s2.a)) % MOD;
return res;
}
ullint p3array[65];
ullint p2array[65];
surd surdarray[65];
void init() {
p2array[1] = 2;
p3array[1] = 3;
surdarray[1].a = 2;
surdarray[1].b = 1;
for(int i = 2; i < 65; i++) {
p2array[i] = modprod( p2array[i-1] , p2array[i-1] );
p3array[i] = modprod( p3array[i-1] , p3array[i-1] );
surdarray[i] = mult(surdarray[i-1],surdarray[i-1]);
}
}
void solve(ullint a) {
if (a == 0 || a % 2 == 1) {
printf("0\n");
return;
}
a = a >> 1;
ullint p2 = 1;
ullint p3 = 1;
surd ps;
ps.a = 1;
ps.b = 0;
int i = 1;
while(a > 0) {
if ( (a & 1) == 1) {
p2 = modprod(p2, p2array[i]);
p3 = modprod(p3, p3array[i]);
ps = mult(ps,surdarray[i]);
}
i++;
a = a >> 1;
}
surd psminus;
psminus.a = modminus(2*ps.a, 3*ps.b);
psminus.b = modminus(2*ps.b, 3*ps.a);
ullint res1 = modprod(MODINV, (2*psminus.a + p3 + 2*p2 + 3) % MOD );
ullint res2 = 2*(2*ps.a + p3 + p2 + 1);
ullint res = modminus(res2, res1);
printf("%llu\n", res);
}
main() {
int T;
scanf("%d", &T);
int i;
init();
for(i = 0; i < T; i++) {
ullint a;
scanf("%llu", &a);
solve(a);
}
}
Stepping Numbers CodeChef Solution in C
“#include <stdio.h>
#define MOD 4294967143U
typedef unsigned int intu;
typedef unsigned long long longu;
longu s1[64],s2[64];
intu pmod(int a, longu n)
{
intu t;
longu s;
s=a;
t=1;
while(n)
{
if (n&1)
t=(s*t)%MOD;
s=(s*s)%MOD;
n>>=1;
}
return t;
}
intu pot(longu n)
{
longu q;
intu t1,t2;
int i;
t1=1;
t2=0;
for(i=0;n;n>>=1,i++)
if (n&1)
{
q=t1;
t1=(s1[i]*t1%MOD+3*(s2[i]*t2%MOD))%MOD;
q=s1[i]*t2%MOD+s2[i]*q%MOD;
if (q>=MOD)
t2=q-MOD;
else
t2=q;
}
return (94LL*t1+144LL*t2)%MOD;
}
int main()
{
int t,i;
longu n;
s1[0]=2;
s2[0]=1;
for(i=1;i<64;i++)
{
s1[i]=(s1[i-1]*s1[i-1]%MOD+3*(s2[i-1]*s2[i-1]%MOD))%MOD;
s2[i]=2*(s1[i-1]*s2[i-1]%MOD)%MOD;
}
scanf("%d",&t);
while(t--)
{
scanf("%llu",&n);
if (n&1)
{
printf("0\n");
continue;
}
n=n-1>>1;
printf("%llu\n",(21+11LL*pmod(2,n+2)+23LL*pmod(3,n+1)+pot(n))%MOD*1789569643%MOD);
}
return 0;
}Stepping Numbers CodeChef Solution Review:
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